Exercise 5.2 · Q5: Differentiate $\frac{\sin(ax+b)}{\cos(cx+d)}$

Basics you need

Solution

1. Name the numerator and the denominator

   $$\begin{aligned} u &= \sin(ax+b), \qquad v = \cos(cx+d), \\ y &= \frac{u}{v}. \end{aligned}$$

   We work at points where $v = \cos(cx+d) \ne 0$⁠, which is exactly where $y$ is defined.

   The last operation in computing $y$ is a division, so the quotient rule is the outermost rule. Naming the two parts keeps the formula $\frac{u'v-uv'}{v^2}$ readable.
   The quotient rule needs the denominator to be non-zero. Basics: derivative rules

2. Differentiate the numerator (chain rule)

   Inner $t = ax + b$ with $\frac{dt}{dx} = a$⁠; outer $\sin t$ with derivative $\cos t$⁠. So

   $$u' = \cos(ax+b)\cdot a = a\cos(ax+b).$$
   This is exactly Question 3. The constant $a$ is the rate at which the inside $ax+b$ changes.
   Chain rule, Theorem 4, Section 5.3.1; $(\sin t)' = \cos t$ from Table 5.3. Exercise 5.2 Q3

3. Differentiate the denominator (chain rule)

   Inner $s = cx + d$ with $\frac{ds}{dx} = c$⁠; outer $\cos s$ with derivative $-\sin s$⁠. So

   $$v' = -\sin(cx+d)\cdot c = -c\sin(cx+d).$$
   Same pattern as the numerator, with cosine in place of sine. The minus sign comes from $(\cos s)' = -\sin s$ and must be carried into the next step.
   Table 5.3: $(\cos s)' = -\sin s$⁠. Lesson: the chain rule

4. Apply the quotient rule

   $$\begin{aligned} \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ &= \frac{a\cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)} \\ &\quad - \frac{\sin(ax+b)\cdot\left(-c\sin(cx+d)\right)}{\cos^2(cx+d)}. \end{aligned}$$

   The two minus signs in the second term multiply to a plus:

   $$\begin{aligned} \frac{dy}{dx} = \frac{1}{\cos^2(cx+d)}\Big[\,&a\cos(ax+b)\cos(cx+d) \\ &+ c\sin(ax+b)\sin(cx+d)\Big]. \end{aligned}$$
   We wrote the quotient rule's numerator over the common denominator as two separate fractions, so that the sign of each piece is visible. The subtraction in $u'v - uv'$ meets the minus sign inside $v'$⁠, and $-(-c) = +c$⁠.
   Quotient rule (3), Section 5.3, valid since $\cos(cx+d) \ne 0$⁠. Basics: derivative rules

5. Turn it into the textbook's form

   Split the fraction into two and simplify each part:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{a\cos(ax+b)}{\cos(cx+d)} \\ &\quad + c\sin(ax+b)\cdot\frac{\sin(cx+d)}{\cos^2(cx+d)} \\ &= a\cos(ax+b)\sec(cx+d) \\ &\quad + c\sin(ax+b)\tan(cx+d)\sec(cx+d). \end{aligned}$$

   In the first fraction one factor $\cos(cx+d)$ cancels (allowed because it is not zero), leaving $\frac{1}{\cos(cx+d)} = \sec(cx+d)$⁠. In the second we split the fraction as

   $$\frac{\sin(cx+d)}{\cos^2(cx+d)} = \frac{\sin(cx+d)}{\cos(cx+d)}\cdot\frac{1}{\cos(cx+d)},$$

   which is $\tan(cx+d)\sec(cx+d)$⁠.

   Both forms of the derivative are correct. The textbook's answer key prints the sec/tan form, so we convert ours into it term by term rather than leave the reader to wonder whether the two agree.
   $\sec\theta = \frac{1}{\cos\theta}$⁠, $\tan\theta = \frac{\sin\theta}{\cos\theta}$⁠. Basics: trigonometric identities

6. Cross-check with the product rule

   Since $\frac{1}{\cos(cx+d)} = \sec(cx+d)$⁠, we can also write $y = \sin(ax+b)\cdot\sec(cx+d)$⁠. The product rule, with $(\sec s)' = \sec s\tan s$ and the inner rate $c$⁠, gives

   $$\begin{aligned} y' &= a\cos(ax+b)\sec(cx+d) \\ &\quad + \sin(ax+b)\cdot c\sec(cx+d)\tan(cx+d), \end{aligned}$$

   the same expression as Step 5.

   Getting the same answer by a different route is strong evidence that no sign or factor was lost. The derivative of $\sec$ used here is derived in Exercise 5.2 Q4, Step 3.
   Product rule (2), Section 5.3: $(uw)' = u'w + uw'$⁠. Basics: derivative rules

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